Maths-
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Easy

Question

I f space u equals tan to the power of negative 1 space end exponent open parentheses fraction numerator x cubed plus y cubed over denominator x minus y end fraction close parentheses space t h e n space x squared fraction numerator partial differential squared u over denominator partial differential x squared end fraction plus 2 x y fraction numerator partial differential squared u over denominator partial differential x partial differential y end fraction plus y squared fraction numerator partial differential squared u over denominator partial differential y squared end fraction equals

  1. open parentheses 2 space C o s 2 u minus 1 close parentheses space S i n space 2 u space
  2. left parenthesis 2 space Cos space space 2 u plus 1 right parenthesis Sin space 2 u
  3. left parenthesis Sin space 2 u minus 1 right parenthesis Cos space 2 u
  4. left parenthesis Sin space 2 u plus 1 right parenthesis Cos space 2 u

The correct answer is: open parentheses 2 space C o s 2 u minus 1 close parentheses space S i n space 2 u space

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